If xy = 3(x + 1) + y; and, x and y are integers, x could be any of

Question

If xy = 3(x + 1) + y; and, x and y are integers, x could be any of the following values except:

A. 7
B. 5
C. 4
D. 3
E. 2

Bunuel · Accepted Answer

You can use the method shown  here . Or do the following:

There are several ways to solve this question through algebraic manipulations. You can use the method shown  here . Or do the following:

Essentially, our goal is to arrive at an equation of the form  x = expression \ with \ y \ only  or  y = expression \ with \ x \ only .

xy = 3(x + 1) + y

xy -y= 3(x + 1)

y(x-1) = 3x + 3

y=\frac{3x + 3}{x-1}

Next, let's attempt to split the fraction to isolate  x  in a single term:

y=\frac{3x -3 + 3 + 3}{x-1}

y=\frac{3x -3 }{x-1}+ \frac{3+3 }{x-1}

x=3+ \frac{6}{x-1}

Since  y  is an integer, for  3+ \frac{6}{x-1}  to be an integer,  \frac{6}{x-1}  must also be an integer. This means that  x-1  must be a divisor of 6. From the options, only if x = 5, x - 1 is not a factor of 6, making it the correct answer.

Answer: B.

Oppenheimer1945 · Answer

xy - 3x - y = 3

xy - 3x - y + 3 = 6

(x - 1)(y - 3) = 6

(1, 6) (-1, -6) (6, 1) (-6, -1) (2, 3) (3, 2) (-2, -3) (-3, -2)

x - 1 = 1, -1, 6, -6, 2, 3, -2, -3

x = 2, 0, 7, -5, 3, 4, -1, -2

KarishmaB · Answer

You have likely forgotten a 'not' in the question.  x  cannot be any of the following values except... 
Or the question is different in some other way. Please check.

Bunuel · Answer

I think it should be "xy = 3(x + 1) + y", instead of "xy = 3(x + 1) + x". In this case x cannot be 5.

KarishmaB · Answer

Ah yes,  Bunuel  ! The question would become logical then.

Lodz697 · Answer

What's the best way to solve a problem like this?

SB2004 · Answer

I plugged in the answer choices (starting with B) into the given equation:

*keeping in mind the given restraints = x/y are integers, I knew to look for the the non-integer value of y

xy = 3(x+1) + y

B) 5y = 3(5+1) + y
5y = 3(6) + y
4y = 18
y is NOT an integer --> B is the answer

If B would not have been the answer, I would've tested D next

Purnank · Answer

Get the equation in terms of y, and plug the values for x.
see if you are getting y as a integer or not. It is best way if you dont want to deal with divisor etc stuffs.

stargaze3 · Answer

My Answer:

xy = 3(x+1)+y

xy-y = 3x+3

y(x-1) = 3x+3

y= (3x+3)/(x-1)

x= 7;

y = 24/6 = 4

x = 5

y = 18/4 ( which is not an integer)

ablatt4 · Answer

xy = 3(x+1)+y
xy-y = 3(x+1)
y(x-1) = 3(x+1)
y = 3(x+1)/(x-1)

I would plug answers in from here and try and solve for the most time efficient way. In more complicated examples isolating x into a single term would be more effective.

ablatt4 · Answer

Solve as much as possible and set up the equation in whatever way is the easiest for you to try plugging in values for x

I recommend xy-y=3+3x

Once you set this up just plug in values based on answer choices until you solve for which isn't an integer

x=5 is going to give you 9/2, the rest will give you intergers.

B

Julioo · Answer

Hi  Bunuel . My question is, as we are not told that x is different from 1, can I still apply that algebraic procedure? How can I be sure I am not dividing by zero in this case when I divide by x-1? Thank you in advance for your help.

egmat · Answer

Julioo  If I can explain - you're absolutely right to check for potential division by zero.

Here's why the algebraic procedure is valid:

Let's test what happens if  x = 1  in the original equation:
 xy = 3(x + 1) + y

Substituting  x = 1 :
 1 \cdot y = 3(1 + 1) + y 
 y = 6 + y 
 0 = 6  ← This is a contradiction!

Key insight:  Since  x = 1  leads to an impossible equation ( 0 = 6 ), we know that  x 
eq 1 . Therefore,  (x - 1) 
eq 0 , and your division by  (x - 1)  is perfectly safe.

Process Diagnosis: 
You correctly identified a potential issue but didn't check whether  x = 1  was actually possible given the constraints. In GMAT problems, when an algebraic manipulation seems necessary, first verify that problematic values lead to contradictions - they usually do!

When you need to divide by an expression: 
 
 First check if setting that expression equal to zero satisfies the original equation 
 If it leads to a contradiction → safe to divide 
 If it works → handle it as a separate case

In GMAT, you'll see this pattern in equation-solving and function problems. The test makers often design problems where the "dangerous" value naturally excludes itself.

Hope this helps!

Bunuel · Answer

The point is x = 1 does not satisfy xy = 3(x + 1) + y for any y, so x cannot be 1.

If xy = 3(x + 1) + y; and, x and y are integers, x could be any of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

­If xy = 3(x + 1) + y; and, x and y are integers, x could be any of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

If xy = 3(x + 1) + y; and, x and y are integers, x could be any of